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E = hf. Got it.

P = dE/dt, right?

Not seeing an amplitude anywhere in the formula, so shouldn't the power rating on the lightbulb only be driven by the color? Is it a problem with the way the lightbulbs physically make light? Maybe making really bright light causes a lot of energy to be lost to heat or something?

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    $\begingroup$ The energy in that formula has nothing to do with the power of the light bulb. It's the energy of individual quanta of light, rather than the macroscopic energy needed to power the bulb. As a matter of engineering, one can produce light bulbs of any power and temperature by varying the thickness and length of the filament. $\endgroup$ – CuriousOne Mar 11 '16 at 21:50
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    $\begingroup$ E=hf refers to the energy and frequency of an individual photon. If you have different flashlights and they are all engineered to emit the same, fixed number of photons per second, then you would be correct in saying that the only way to change the power rating of a flashlight (or light bulb) would be to change the color (i.e., the energy of the individual photons). $\endgroup$ – Samuel Weir Mar 11 '16 at 21:50
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Your first equation describes the energy of a single photon, which only depends on frequency.Though the intensity or brightness of light depends on the number of photons as well which is proportional to the square of the amplitude.

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You are confusing two different concept. Both of them describe light but they are structured differently.

  • Your title refers to a "wave". A "wave" is a classical description of light in terms of Maxwell's equations. The energy density of a classical light wave depends on the amplitude of the wave and not on the frequency.

  • The body of your question uses the expression $E = hf$, which is part of a quantum description of light in terms of photon (light particles, more or less) In particular this gives the energy of a photon in terms of it's frequency.

Either description is valid, but you have to use them on their own terms and not try to mix them willy-nilly. (Well, once you know them very well you'll find that you can get away with mixing them sometimes; but sometimes is slippery.)

The reason amplitude doesn't appear in the quantum formula you are asking about is because that expression applies to one photon, and the strength of the light is related to the number of photons present.


The advice I give beginners about light is: pick one description and learn it well on it's own merits ("well" being defined by being able to calculate confidently using all the concepts in that description), then learn the other one well (same criteria), and finally start thinking about the relationship between them when you are able compare the results of non-trivial calculation between frameworks.

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There are three elements to an answer to this question. The first is, as mentioned above by others, the power is dependent on both the energy of the individual photons and the number of photons that are emitted every second.

Secondly the light from different light bulbs actually emit a spectrum of different frequencies. For the old incandescent bulbs, only a small percent of the emissions were visible, so most of the power was used to create non visible light.

Which brings us to the third factor. Our eyes are sensitive to only a small range of frequencies, and even then our the amount we see as the same brightness varies with frequency. This means that some lights could seem dimmer to us, even if they are brighter.

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protected by Qmechanic Mar 11 '16 at 22:48

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